Persistent Bugger 实践练习

最新推荐文章于 2022-10-26 10:00:00 发布

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本文介绍了一个有趣的数学问题，即计算一个数的乘积持久度。通过将一个数的各位数字相乘，再将结果的各位数字相乘，直至得到一位数，过程中需要进行的乘法操作的次数即为该数的乘积持久度。文章提供了Python实现的代码示例。

Write a function, persistence, that takes in a positive parameter num and returns its multiplicative persistence, which is the number of times you must multiply the digits in num until you reach a single digit.
编写一个函数，任意一个整数，拆分成个位数然后相乘，得到的乘积再次相乘，直到乘积为单数为止，函数返回拆分次数。

For example: 举例：

persistence(39) => 3  # Because 3*9 = 27, 2*7 = 14, 1*4=4
                       # and 4 has only one digit.

persistence(999) => 4 # Because 9*9*9 = 729, 7*2*9 = 126,
                       # 1*2*6 = 12, and finally 1*2 = 2.

persistence(4) => 0   # Because 4 is already a one-digit number.

练习源码：

def persistence(n):
    sum = n
    total_times = 0
    while sum > 9:
        temp = 1
        for i in str(sum):
            temp = temp*int(i)
        sum = temp
        total_times += 1
    return total_times